Students:  I need to go back a few lectures and review an important point.  The following information has been added to Lecture 3. 

A note on life cycles and the time of recombination:  In higher organisms, the next generation begins at the time of fertilization.  Two haploid gametes fuse to form the diploid zygote.  The resulting single diploid cell then undergoes mitosis to produce two identical diploid cells. Genetic information is passed on to the next two cells and the resulting two cells are genetically identical (barring mutation).   This series of mitotic events continues to occur to produce virtually all of the cells of the organism.  A few diploid cells will give rise to the gametes that  fuse to form the next generation.  In the process of gamete formation, diploids cells undergo meiosis to form haploid cells.  It is during meiosis that recombination occurs.  For example in our fly experiment above, the results of fertilization between the two homozygous parents (lz^g /lz^g   and lz ^s/lz ^s )  is lz ^g/lz ^s .   All diploid cells in this heterozygous individual are of that genotype.  BUT  a few resulting haploid gametes arising from this heterozygote can be wildtype because of recombination.  Because we cannot look at gametes to score this phenotype, we must wait until the next generation to score the eye phenotype.

End of note

    The following diploid genotypes were made and induction of enzymes were studied. For brevity and in the tradition of genetics, only the mutant genes are listed. Wildtype forms of the other relevant genes are present, ie, the diploid containing wildtype forms of all genes except the i gene is written, i⁺/i^-.

                                                                                        Enzymes present

Genotype:                                                    No inducer            Plus inducer

wildtype (haploids)                                                 -                                 +

       i-                                                                     +                                +

       p-                                                                    -                                  -

      O^c                                                   +                                 +
 

Now the diploids:

      i⁺/i^-                                                                   -                                +

      p⁺ /p-                                                               -                                +

      O⁺ /O^c                                                             +                                +
 

        Note that i and p mutants above are recessive while the O^c mutant is dominant.
 

        Definitive evidence concerning the exact roles of the regulatory genes came from placement of regulatory mutants in cis or trans arrangement with structural gene mutations. The three combinations of regulatory mutants with a structural gene mutation (in this case z) were made in E. coli cells.

First:
 

           i^-                z^-           (cis)                         and          i⁺                          z^-                trans
         i⁺                  z⁺                                                           i^-                 z⁺
 

    These had identical (and wild type) phenotypes: (no B-gal in the absence of inducer and activity in the presence of inducer).

Second:

             p^-              z^-                     (cis)                        and           p^-                    z⁺      (trans)
           p⁺                z⁺                                                                     p⁺                    z^-
 

    Here the phenotypes are different. Whereas the second one, trans, showed the mutant phenotype (no enzyme in the presence of inducer), the first genotype gave the wild type phenotype. We will return to this genotype and the meaning of the cis trans test.
 

Third:
       O^c              z^-                 (cis)             and            O^C              z⁺ (trans)
       O⁺                  z⁺                                                       O⁺         z^-
 
 

The phenotype is again different. Whereas the first arrangement, cis is wild type (enzyme only in the presence of inducer), the second one produces enzyme in the absence of inducer (ie the  phenotype of O^c mutants.
 

    Important: Note in the second and third cases above that expression of the functional z gene is determined by which allele of the affected gene is  cis to it. In the case of the heterozygote with the p gene, the functional or z⁺ gene is expressed in wild type manner only when it is cis to the p⁺gene. When it is cis to the mutant p^- allele, it is not expressed (ie the mutant phenotype.). The same is true in the third case involving the O mutant. The expression pattern of the functional z gene is determined strictly by the cis allele of the gene in question.
 

The fact that p and O acted only on the structural genes located on the same DNA fragment led to the idea that these genes (i and O) did not encode a protein product. In contrast, the functional i gene, i⁺ can act both in cis and in trans. Hence it was concluded that it produces a protein product.

    How do we know that there is one message but three separate proteins?

     Polar mutants provided definitive insight. Early in the investigations, exceptional mutants were found that mapped to the z gene and not only knocked out B-gal but also greatly reduced expression of the other two enzymes. Likewise, there were exceptional y mutants that reduced permease and transacetylase (but not B-gal). However, none of the a mutants affected the two other enzymes. The facts that exceptional z mutants affected y and a, that exceptional y mutants affected a but not z and that no a mutants that these effects gave rise to the concept of polarity.

    Significantly, many of these polar mutants were suppressible by non-sense suppressors. Non-sense suppressors result from base substitutions in the DNA corresponding to the anticodon of a tRNA that  inserts an amino acid into the protein.  The mutant anti-codon now recognizes a stop or chain termination codon, inserting an amino acid at this position.

    Quite significantly, suppression of these non-sense mutations also led to suppression of polarity. The only simple logical conclusion to draw from this is that the translation of these three genes is done in concert. If the transcripts for these three genes were separate and independent, then one would not expect polar mutants. (I am totally unaware of any polar mutants in higher plants. )
 

 Think about the definition of the cistron and the phenotype of z^- and p^- mutants when present in the cis and trans arrangement. If mutants give rise to the mutant phenotype when present in the trans arrangement but wild type when present in the cis arrangement, they are said to be in the same cistron. By definition then, mutants of p and z would belong to the same cistron.

Therefore, the cistron corresponds to a structural gene plus any cis-dominant regulatory genes.

 
1.)  Non-allelic non-complementation.   See notes from lecture 6.  Remember, this is a rare event!!! 
 

2.) Recall the discussion above of polar mutants. Suppose we have a polar mutant in z and a typical mutant in y. Suppose our phenotype is the presence or absence of the y product, permase (in the presence of inducer). Perform the cis/trans test with these two mutants. We would conclude that they are in the same cistron, yet they are in two separate genes. What is different here, compared to higher organisms is that the two genes are in the same operon or the same transcribed region. If two genes are not transcribed together in one primary transcript, then we would not have this case. Hence these cases are rare.

3. ) Intracistronic or interallelic complementation. It has been noted in some systems that when an enzyme is composed of two or more identical subunits, two different, and mutant subunits can sometimes interact to produce an enzyme with some enzymic activity. For example, suppose we have an enzyme that is what we call a homodimer (homo meaning same, dimer meaning two) and we call the subunit A. Hence the active enzyme would be AA. Suppose we have two mutant alleles of this gene. These produce subunits we can term a1-X and a1-Y. By definition, dimers of the composition a1-X/a1-X and a1-Y/a1-Y would be nonfunctional or mutant. However, in rare cases, the heterozygote, which produces at a frequency of 50% dimers of the constitution a1-X/a1-Y have a wildtype phenotype. This is due to the fact that the a1-X/a1-Y dimer has partial activity. If the level of activity is high enough to throw it in to the "wildtype" threshold, then the wildtype phenotype occurs. This is seen even when they are in the same structural gene. There are several things involved here: the enzyme must be made up of 2 or more identical subunits, the mutant proteins must interact in a peculiar way and the difference in enzymic activity conditioned by the enzyme from the heterozygote must give rise to a change in phenotype. All of this makes this a rare or exceptional phenomenon. Hence these cases are rare.


   Rare means that it is rare to find two alleles that exhibit this phenomenon.  BUT, when one has two such alleles, ONE GETS THE WILDTYPE PHENOTYPE EVERYTIME THESE TWO ALLELES ARE CROSSED!!!    In other words, if we cross two particular mutants and we see a few rare wildtype progeny among a vast majority of mutant progeny, the rare wildtypes are NOT due to intracistronic complementation.

Epistasis

A phenomenon related to all of this is termed epistasis.   Epistasis refers to suppression of the mutant phenotype determined by recessive alleles of one gene by the presence of homozygosity of a mutant gene at another locus.   A couple of examples:

A cross between a  plant containing a recessive a1 mutant allele in homozygous condition (but functional homozygous genes of all other genes required for pigment synthesis) with a  plant containing a bz1 mutant allele in homozygous condition (but functional homozygous genes of all other genes required for pigment synthesis) produces plants with the wildtype purple pigment.  This is diagramed below.  Note that only the contrasting alleles at just two (A1 and Bz1) of the many genes required for pigment synthesis are listed:

     a1/a1;  Bz1/Bz1 (colorless)      X    A1/A1;  bz1/bz1  (bronze)

      F₁   is A1/a1;  Bz1/bz1  (purple)

      F₂     is   9     A1/-;  Bz1/-  (purple)

                   3     A1/-   bz1/bz1 (bronze)

                   4     a1/a1;  -/-    (colorless)

            Note that this ratio 9:3: 4 is a modification of the 9:3:3:1 ratio we are accustomed to. This is explained by the fact that if a plant lacks A1 function, there is no color regardless of the genotype at the Bz1 locus.  The A1 locus is said to be epistatic to Bz1.  

      Can epistasis always be used to order steps in a pathway?  Absolutely and categorically not!!!!!

            9/16 plump

            3/16 mildly shrunken

            4/16 severely shrunken

            If we apply the logic that the epistatic locus is upstream, as in the case of A1 and Bz1 above, then Sh2 would be upstream to Sh1.  The biochemistry however showed just the opposite:

            The Sh2 gene encodes one subunit of an enzyme that comes much later in the starch biosynthetic pathway, namely adenosine diphosphoglucose pyrophosphorylase.  This enzyme synthesizes ADP-glucose from glucose-1- phosphate and ATP.  ADP-glucose is the glucose donor for starch production.  While there is a second form of this enzyme, the relative contribution to total ADP-glucose pyrophosphorylase activity is much less, compared to the sucrose synthase case above.